Students should refer to ICSE Class 10 Physics Question Paper solved Set E given below which will help them to prepare for the upcoming ICSE Physics exams. Students should read ICSE Physics Class 10 Books to make sure they are completely prepared and should also refer to ICSE Class 10 Physics Solutions to understand all questions and their answers.
ICSE Class 10 Physics Question Paper solved Set E
Answers to this paper must be written on the paper provided separately.
You will not be allowed to write during the first 15 minutes.
This time is to be spent in reading the Question Paper.
The time given at the head of this paper is the time allowed for writing the answers.
Section I is compulsory. Attempt any four questions from Section II.
The intended marks for questions or parts of questions are given in brackets [ ].
ICSE Class 10 Physics Question Paper solved Set E
SECTION – I (40 Marks)
(Attempt all Questions)
Question 1
(a)(i) Give an example of a non contact force which is always of attractive nature.
(ii) How does the magnitude of this non contact force on the two bodies depend on the distance of separation between them? [2]
(b) A boy weighing 40 kgf climbs up a stair of 30 steps each 20 cm high in 4 minutes and a girl weighing 30 kgf does the same in 3 minutes. Compare:
(i) The work done by them.
(ii) The power developed by them. [2]
(c) With reference to the terms Mechanical Advantage, Velocity Ratio and efficiency of a machine, name and define the term that will not change for machine of a given design. [2]
(d) Calculate the mass of ice required to lower the temperature of 300 g of water 40oC to water at 0oC. [2]
(Specific latent heat of ice = 336 J/g, Specific heat capacity of water = 4.2J/goC)
(e) What do you understand by the following statements:
(i) The heat capacity of the body is 60JK-1.
(ii) The specific heat capacity of lead is 130 Jkg-1K-1. [2]
Answer:
(a)(i) The example of non-contact force is gravitational force. It is always attractive in nature.
(ii) The magnitude of non-contact forces acting on two bodies depends on the distance of separation between them. The magnitude of force decreases with an increase in separation and increases as the separation decreases. It varies inversely as the square of the distance of separation.
(b) Given:
Force of gravity of the boy, Fb = 40 kgf
Time taken by him, tb = 4 minutes = 4 × 60 s = 240 s
Force of gravity of the girl, Fg = = 30 kgf
Time taken by her, tg = 3 minutes = 3 × 60 s = 180 s
Distance covered by both in 30 steps is
D = 30 × 20 = 600 cm
While climbing, both have to do work against the force of gravity.
(i) Work done by the boy in climbing the stairs:
Wb = F × D = 40 kgf × 600 cm
Wb = 24000 J
Work done by the girl in climbing the stairs:
Wg = F × D = 30 kgf × 600 cm
Wg = 18000 J
→ Wb/Wg = 24000J/18000J
∴ Wb/Wg = 4/3
(ii) Power developed = Work done/Time taken
For the boy :
Power Developed =24000J / 240 s = 100W
For the girl :
Power Developed =18000J/180s = 100W
Thus, power developed by them is 1 : 1
Or
(i) Work done by the boy = 40 × 10 × 6 = 2400 J
Work done by the girl = 30 × 10 × 6 = 9800 J
∴ Ratio = 2400 : 1800 = 4 : 3
(ii) Power of boy =2400/4 × 60 = 10 watt
Power of girl =2400/3 × 60 = 10 watt
∴ Ratio = 1 : 1
(c) Velocity ratio.
It is the ratio of the velocity of effort to the velocity of load.
V.R. = VE/VL .
Velocity ratio does not change.
(d) Let m be the mass of the ice to be added.
Heat energy required to melt to lower the temperature is = mL = m × 336
Heat energy imparted by the water in fall of its temperature from 40°C to 0°C = mass of the water × specific heat capacity × fall in temperature.
= 300 × 4.2 × 40°C
If there is no loss of heat,
m × 336 J/g = 300 g × 4.2 J/g°C × 40°C
∴ m = 300 × 4.2 × 40 / 336
∴ m = 150g
(e) (i) Heat capacity is the amount of heat required to raise the temperature of a body by 1°C or 1 K.
Thus, 60 JK–1 of energy is required to raise the temperature of the given body by 1 K.
(ii) Specific heat capacity is the amount of heat energy required to raise the temperature of unit mass of a substance through 1°C or 1K. Thus, 130 JKg–1 K–1 of heat energy required to raise the temperature of unit mass of lead through 1 K.
Question 2
(a) State two factors upon which the heat absorbed by a body depends. [2]
(b) A boy uses blue colour of light to find the refractive index of glass. He then repeats the experiment using red colour of light. Will the refractive index be the same or different in the two cases? Give a reason to support your answer. [2]
(c) Copy the diagram given below and complete the path of the light ray till it emerges out of the prism. The critical angle of glass is 42°. In your diagram mark the angles wherever necessary. [2]
(d) State the dependence of angle of deviation:
(i) On the refractive index of the material of the prism.
(ii) On the wavelength of light [2]
(e) The ratio of amplitude of two waves is 3:4. What is the ratio of their:
(i) loudness?
(ii) Frequencies? [2]
Answer :
(a) Heat absorbed by a body depends on the mass of the body, change in temperature of the body and the specific heat capacity of the body.
(b) The refractive index will be different in both cases.
Refractive index of glass is different for different colours. The speed of blue light is less than the speed of red light. So, the wavelength of blue light is less than that of red light. Thus, red light would deviate less than blue light because of difference in wavelength.
(c) The diagram is as shown :
(d)(i) Angle of deviation is directly proportional to the refractive index of the material of prism. For a given angle of incidence, the prism with higher refractive index produces a greater deviation than the prism which has a lower refractive index. Thus, the angle of deviation increases with an increase in the refractive index of the medium.
(ii) Angle of deviation is inversely propotional to the wavelength of the light used. The angle of deviation decreases with an increase in the wavelength of light. Thus, a prism deviates violet light the most and red light the least.
(e) (i) Let a1 and a2 be the amplitudes and I1 and I2 be the intensities of the two waves.
(ii) Frequency is the number of waves formed per second. It only depends on time period. Thus, the ratio of their frequencies is 1:1.
Question 3
(a) State two ways by which the frequency of transverse vibrations of a stretch string can be increased. [2]
(b) What is meant by noise pollution? Name one source of sound causing noise pollution. [2]
(c) The V-I graph for a series combination and for a parallel combination of two resistors is shown in the figure below. Which of the two A or B. represents the parallel combination? Give reasons for your answer.
(d) A music system draws a current of 400 mA when connected to a 12 V battery. [2]
(i) What is the resistance of the music system?
(ii) The music system if left playing for several hours and finally the battery voltage drops to 320 mA and the music system stops playing when the current.
(e) Calculate the quantity of heat produced in a 20Ω resistor carrying 2.5 A current in 5 minutes. [2]
Answer :
(a) The frequency of transverse vibration is given by
where l = length of the vibrating string
T = tension in the string
m = mass per unit length of the string
Therefore, the frequency of transverse vibration of a stretched string can be increased by
(i) decreasing the length of the string
(ii) decreasing the mass per unit length of the string
(iii) increasing the tension T in the string
(b) The disturbance produced in the environment because of undesirable loud and harsh sound of level more than 120 dB from the various sources such as a loudspeaker and moving vehicles is called noise pollution.
Ex : Honking of vehicles in traffic jams.
(c) For the same change in I, change in V is less for the straight line A than for the straight line B (i.e. the straight line A is less steep than B). The straight line A represents small resistance, while the straight line B represents more resistance. The equivalent resistance is less in a parallel combination than in a series combination. So, line A represents a parallel combination.
(d) (i) Given : I = 400 mA = 400 × 10–3 A
V = 12 V
V = IR
∴ R = V/I = 12V/400×10-3 A
∴ R = 30Ω
(ii) Current drops to I = 320 mA = 320 × 10–3 A
The music stops playing at
V = IR
= 320 × 10–3 × 30
∴ V = 9.6 V
(e) Given : R = 20Ω , I = 2.5 A
t = 5 minutes = 5 × 60 = 300 s
Quantity of heat produced is given as
H = I2Rt
= (2.5)2 × 20 × 300
∴ H = 37500 J
Question 4
(a) State the characteristics required of a good thermion emitter. [2]
(b) An element ZSA decays to 85R222 after emitting 2 α particles and 1 β particle. Find the atomic number and atomic mass of the element S. [2]
(c) A radioactive substance is oxidized. Will there be any change in the nature of its radioactivity? Give a reason for your answer. [2]
(d) State the characteristics required in a material to be used as an effective fuse wire. [2]
(e) Which coil of a step up transformer is made thicker and why? [2]
Answer :
(a) Two characteristic which a thermionic emitter should possess are
1. The work function of the substance should be low such that the electrons are emitted from it even when it is not heated to a very high temperature.
2. The melting point of the substance should be quite high so that it may not melt when it is heated to the temperature required for thermionic emission.
(b) The decay will follow the following sequence
Therefore, we have
Z – 3 = 85 or Z = 85 + 3 = 88
And A – 8 = 222 or A = 222 + 8 = 230
Z = 88, A = 230
(c) No, radioactivity is not affected by any physical process.
(d) The material should have high resistivity and low melting point.
(e) Primary coil, as it has lesser number of turns.
ICSE Class 10 Physics Question Paper solved Set E
SECTION II (40 Marks)
(Attempt any four questions from this Section)
Question 5
(a) A stone of mass ‘m’ is rotated in a circular path with a uniform speed by tying a strong string with the help of your hand. Answer the following questions : (3)
(i) Is the stone moving with a uniform or variable speed ?
(ii) Is the stone moving with a uniform acceleration ? In which direction does the acceleration act?
(iii) What kind of force acts on the hand and state its direction ?
(b) From the diagram given below. answer the question that follow : (3)
(i) What kind of pulleys are A and B ?
(ii) State the purpose of pulley B.
(iii) What effort has to be applied at C just raise the load L = 20 kgf ?
(Neglect the weight of pulley A and friction)
(c) (i) An effort is applied on the bigger wheel of a gear having 32 teeth. It is used to turn a wheel of 8 teeth. Where it is used.
(ii) A pulley system has three pulleys. A load of 120 N is overcome by applying an effort of 50 N.
Calculate the Mechanical Advantage and Efficiency of this system. (4)
Answer :
(a)(i) The stone is moving with uniform speed as given in the question.
(ii) Although the stone is rotating with uniform speed, its direction keeps on acceleration. The direction of acceleration is towards the centre of the circular path.
(iii) Centrifugal force acting radially outwards.
(b)(i) The pulley A is a single movable pulley and B is a single fixed pulley.
(ii) So as to apply the effort in a convinient direction that is vertically downwards.
(iii) Distance moved by load dL = x / 2.
Distance moved by the effort dE = x
Load = 20 kgf, Effort = ?
Now, Load × dL = Effort × dE
20 × x / 2 = Effort × x
or Effort = 10 kgf
(c) (i) The number of teeth in the driving wheel is nB = 32 and that in the driven wheel is nA = 8. Hence, this system is used to obtain gain in speed.
Gain in speed = nB/nA = 32/8 = 4
(ii) Load = 120 N; E = 50 N; n = 3
Mechanical advantage of pulley system is
M.A. =L/E =120/50 = 2.4
The efficiency of the system is
η = M.A/V.A = 2.4/n = 2.4/3 =0.8=80%
Question 6.
(a)(i) What is the principle of method of mixtures ? (3)
(ii) What is the other name given to it ?
(iii) Name the law on which the principle is based
(b) Some ice is heated at a constant rate, and its temperature is recorded after every few seconds, till steam is formed at 100°C. Draw a temperature time graph to represent the change. Label the two phase changes in your graph. (3)
(c) A copper vessel of mass 100 g contains 150 g of water at 50°C. How much ice is needed to cool it to 5°C ? (4)
Given : Specific heat capacity of copper = 0.4 Jg–1 °C–1
Specific heat capacity of water = 4.2 Jg–1 °C–1
Specific latent heat of fusion ice = 336 Jg–1
Answer :
(a)(i) The principle of method of mixture says that the heat lost by a hot body is equal to the heat gained by a cold body provided no heat is last to the surroundings.
(ii) The other name given to the principle of mixture is the principle of calorimetry.
(iii) The principle of mixture is based on the law of conservation of energy.
(b) The figure for phase change is shown below :
(c) Heat energy lost by the vessel and water contained in it in cooling the water from 50°C to 5°C is used in heating ice to melt it and then to raise its temperature from 0°C to 5°C.
Now, heat energy lost by the copper vessel is
Question 7.
(a)(i) Write a relationship between angle of incidence and angle of refractions for a given pair of media.
(ii) When a ray of light enters from one medium to another having different optical densities it bends. Why does this pehnomenon occur ?
(iii) Write one conditions where it does not bend when entering a medium of different optical density.(3)
(b) A lens produces a virtual image between the object and the lens. (3)
(i) Name the lens.
(ii) Draw a ray diagram to show the formation of this image.
(c) What do you understand by the term ‘Scattering of light’ ? Which colour of white light is scattered the least and why ? (4)
Answer :
(a)(i) The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media called the refractive index = sini / sinr =μ
(ii) When a ray of light enters from one medium to another with different optical densities, it bends because there is a change in the speed of light in the two medium.
(iii) A ray of light passing from one medium to another does not bend when incident normally at the surface of the second medium.
(b) (i) The image formed by the lens is virtual and between the object and the lens. Hence, the lens used is a concave lens.
(ii) The following image is of formation of the above image :
(c) Scattering is the process of absorption and then re-emission of light energy without the change in the wavelength.
The red colour of the white light is scattered the least because scattering of light depends inversely upon the four power of wavelength. As red colour has the maximum wavelength in the visible region, therefore, it scattered the least.
Question 8.
(a) (i) Name the waves used for echo depth sounding.
(ii) Give one reason for their use for the above purpose.
(iii) Why are the waves mentioned by you not audible to us ? (3)
(b) (i) What is an echo
(ii) State two conditions for an echo to take place. (3)
(c) (i) Name the phenomenon involved in tuning a radio set to a particular station.
(ii) Define the phenomenon named by you in part (i) above.
(iii) What do you understand by loudness of sound ?
(iv) In which units is the loudness of sound measured ?
Answer :
(a) (i) The waves used for echo depth sounding are ultrasonic waves.
(ii) Ultrasonic waves are used for echo depth ranging because they can travel undeviated through a long distance.
(iii) Ultrasonic waves have frequency larger than 20000 Hz. Hence, these waves are not audible to us as the audible range for the human ear is 20 Hz to 20000 Hz.
(b) (i) The sound heard after reflection from a distant obstacle after the original sound has ceased is called an echo.
(ii) The conditions for an echo to take place are
a. The minimum distance between the source of sound and the reflector in air must be at least present 16 to 18 m.
b. The size of the reflector must be large enough as compared to the wavelength of sound wave.
(c) (i) The phenomenon involved in tuning a radio set to a particular station is called resonance.
(ii) Resonance : When the frequency of an externally applied periodic force on a body is equal to its natural frequency, the body readily begins to vibrate with an increased amplitude. This phenomenon is known as resonance.
(iii) Loudness is the property by virtue of which a loud sound can be distinguished from a faint one, both having the same pitch and quality.
(iv) The loudness of sound is measured in phon / db / bel.
Question 9.
(a) (i) Which particles are responsible for current in conductors ?
(ii) To which wire of a cable in a power circuit should the metal case of a geyser be connected ? (3)
(b) (i) Name the transformer used in the power transmitting station of a power plant.
(ii) What type of current is transmitted from the power station ?
(iii) At what voltage is this current available to our household ? (3)
(c) A battery of emf 12 V and internal resistance 2Ω is connected with two resistors A and B of resistance 4Ω and 6Ω respectively joined in series. (4)
Find :
(i) Current in the circuit
(ii) The terminal voltage of the cell.
(iii) The potential difference across 6Ω Resistor.
(iv) Electrical energy spent per minute in 4Ω resistor.
Answer :
(a)(i) Electrons are responsible for current in conductors.
(ii) The metal case of a geyser should be connected to the Earth wire.
(iii) The fuse should always be connected to the live wire.
(b)(i) A step-up transformer is used in the power transmitting station of a power plant.
(ii) An alternating current is transmitted from the power station.
(iii) The current is available to our household at a voltage of 220 V.
(c) E = 12 V ; r1 = 2Ω; RA = 4Ω; RB = 6Ω
(i) The current in the circuit is
I = E / Rtotal = E / R1 + RA + RB
∴ I = 12 / 2+4+6 = 1A
(ii) The terminal voltage of the cell is
T.V. = IR
T.V. = 1 × (6 + 4)
T.V. = 10 V
(iii) The potential difference across the 6Ω resistor is
VB = IRB
∴ VB = 1 × 6 = 6 V
(iv) The electrical energy spent per minute (= 60s) is
W = I2Rt
W = 12 × 4 × 60 = 240 J
Question 10.
(a) Arrange a , b , and g rays in ascending order with respect to their (3)
(i) Penetrating power.
(ii) Ionising power.
(iii) Biological effect.
(b)(i) In a cathode ray tube what is the function of anode ?
(ii) State the energy conversion taking place in a cathode ray tube.
(iii) Write one use of cathode ray tube. (3)
(c) (i) Represent the change in the nucleus of a radioactive element when a b particle is emitted.
(ii) What is the name given to elements with same mass number and different atomic number.
(iii) Under which conditions does the nucleus of an atom tend to radioactive ? (3)
Answer :
(a) (i) Penetrating power : α < β < γ
(ii) Ionising power : γ < β < α
(iii) Biological effect = α < β < γ
(b)(i) In a cathode ray tube, the anode accelerates the electrons and also focuses them in a fine energetic beam.
(ii) In a cathode ray tube, electrical energy is converted into heat and than light energy.
(iii) A cathode ray tube is used to investigate the wave form of an unknown alternating potential by applying it on the Y-plates and a known periodic time-based potential on the X-plates or picture tube of T.V.
(c) (i) In an unstable nucleus, the neutron is changed into a proton by emitting a beta particle. This is represented as
(ii) Elements with the same mass number but different atomic numbers are called isobars.
(iii) The nucleus of an atom tends to be radioactive when
— nucleus has more mass
— nucleus has excess energy
— neutrons proton ratio becomes more than 1.3 to 1.5.